无法在声明的路径中找到文件

时间:2013-09-27 16:39:11

标签: c# winforms

C#,WFA,使用.NET 4.5平台,

我删除了textBox1(名字),textBox2(姓氏),pictureBox1(员工照片),button2(浏览)和button1(保存)以插入新员工。

button2 - >必须浏览图片并在pictureBox1中显示,

button1 - >必须保存图像,该图像由button2浏览并由pictureBox1显示,to this table in localhost.

运行程序后,I GET THIS ERROR (file could not be found.)

(虽然我单独浏览时没有任何错误)

我只想要包含修复此WFA的代码的答案。只是想确定我对此非常清楚。

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;

using System.IO;
using System.Data.SqlClient;


namespace WindowsFormsApplication1
{
    public partial class Form1 : Form
    {

        SqlConnection cnn = new SqlConnection("Initial Catalog=randomcompany;Data Source=localhost;Integrated Security=SSPI;");

        public Form1()
        {
            InitializeComponent();
        }

        private void Form1_Load(object sender, EventArgs e)
        {

        }

        private void button2_Click(object sender, EventArgs e) //Browse button
        {
            try
            {
                OpenFileDialog dlg = new OpenFileDialog();
                dlg.Filter = "Images (*.BMP;*.JPG;*.GIF)|*.BMP;*.JPG;*.GIF|" + "All files (*.*)|*.*";
                dlg.Title = "Select Employee Picture";
                if (dlg.ShowDialog() == DialogResult.OK)
                {
                    pictureBox1.Image = System.Drawing.Image.FromFile(dlg.FileName);
                    pictureBox1.SizeMode = PictureBoxSizeMode.StretchImage;
                }
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }
        }

        private void button1_Click(object sender, EventArgs e) //Save button
        {

            try
            {
                cnn.Open();
                string path = pictureBox1.Image.ToString();
                Byte[] imagedata = File.ReadAllBytes(path);
                SqlCommand cmd = new SqlCommand("INSERT INTO Employees (EmployeeFirstname, EmployeeLastname, EmployeePhoto) VALUES (@item1,@item2,@img", cnn);
                cmd.Parameters.AddWithValue("@item1", textBox1.Text);
                cmd.Parameters.AddWithValue("@item2", textBox2.Text);
                cmd.Parameters.AddWithValue("@img", imagedata);
                cmd.ExecuteNonQuery();
                cnn.Close();
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }

        }

    }
}

3 个答案:

答案 0 :(得分:4)

当您尝试从图像中获取图像文件路径时:

string path = pictureBox1.Image.ToString();

你实际上得到的是Image type(System.Drawing.Bitmap)的名称

在对话框中选择后,需要保留图像文件的路径。你可以这样做:

pictureBox1.Image = System.Drawing.Image.FromFile(dlg.FileName);
pictureBox1.Tag = dlg.FileName;

然后你需要读这个名字:

string path = pictureBox1.Tag as string;

答案 1 :(得分:3)

问题在于:

string path = pictureBox1.Image.ToString();

不会返回路径。在类字段中获取路径时存储路径。我们将其命名为_path

private string _path;

然后在获取文件名时设置它:

_path = dlg.FileName;

然后在这里使用它:

Byte[] imagedata = File.ReadAllBytes(_path);

以下是修改后的完整代码:

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;

using System.IO;
using System.Data.SqlClient;


namespace WindowsFormsApplication1
{
    public partial class Form1 : Form
    {
        private string _path;

        SqlConnection cnn = new SqlConnection("Initial Catalog=randomcompany;Data Source=localhost;Integrated Security=SSPI;");

        public Form1()
        {
            InitializeComponent();
        }

        private void Form1_Load(object sender, EventArgs e)
        {

        }

        private void button2_Click(object sender, EventArgs e) //Browse button
        {
            try
            {
                OpenFileDialog dlg = new OpenFileDialog();
                dlg.Filter = "Images (*.BMP;*.JPG;*.GIF)|*.BMP;*.JPG;*.GIF|" + "All files (*.*)|*.*";
                dlg.Title = "Select Employee Picture";
                if (dlg.ShowDialog() == DialogResult.OK)
                {
                    pictureBox1.Image = System.Drawing.Image.FromFile(dlg.FileName);
                    _path = dlg.FileName;
                    pictureBox1.SizeMode = PictureBoxSizeMode.StretchImage;
                }
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }
        }

        private void button1_Click(object sender, EventArgs e) //Save button
        {

            try
            {
                cnn.Open();
                Byte[] imagedata = File.ReadAllBytes(_path);
                SqlCommand cmd = new SqlCommand("INSERT INTO Employees (EmployeeFirstname, EmployeeLastname, EmployeePhoto) VALUES (@item1,@item2,@img", cnn);
                cmd.Parameters.AddWithValue("@item1", textBox1.Text);
                cmd.Parameters.AddWithValue("@item2", textBox2.Text);
                cmd.Parameters.AddWithValue("@img", imagedata);
                cmd.ExecuteNonQuery();
                cnn.Close();
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }

        }

    }
}

答案 2 :(得分:1)

您使用了错误的属性来获取图像路径。你应该 图片框的PictureBox.ImageLocation属性可以获取图像的确切位置。

修改此部分

private void button1_Click(object sender, EventArgs e) //Save button
    {

        try
        {
            cnn.Open();
            string path = pictureBox1.ImageLocation; // this will work
            string path = pictureBox1.Image.ToString(); // here error comes
            Byte[] imagedata = File.ReadAllBytes(path);
            SqlCommand cmd = new SqlCommand("INSERT INTO Employees (EmployeeFirstname, EmployeeLastname, EmployeePhoto) VALUES (@item1,@item2,@img", cnn);
            cmd.Parameters.AddWithValue("@item1", textBox1.Text);
            cmd.Parameters.AddWithValue("@item2", textBox2.Text);
            cmd.Parameters.AddWithValue("@img", imagedata);
            cmd.ExecuteNonQuery();
            cnn.Close();
        }
        catch (Exception ex)
        {
            MessageBox.Show(ex.Message);
        }

    }